Suppose we have a plane $P$ defined by the transformation $T$ for $-v < u < 4$ and $-2 < v < 1$. $T(u, v) = (4u - 2v, u + v, 2u + 3v)$ What is the surface area of $P$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{-2}^1 \int_{-v}^4 2 \sqrt{67} \, du \, dv$ (Choice B) B $ \int_{-2}^1 \int_{-v}^4 \sqrt{293} \, du \, dv$ (Choice C) C $ \int_{-2}^1 \int_{-v}^4 7 \sqrt{5} \, du \, dv$ (Choice D) D $ \int_{-2}^1 \int_{-v}^4 2 \sqrt{307} \, du \, dv$
Solution: Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_{-2}^1 \int_{-v}^4 |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = \sqrt{293}$ [Calculation] In conclusion, the surface area of $P$ is: $ \int_{-2}^1 \int_{-v}^4 \sqrt{293} \, du \, dv$